1. Derivatives

Need derivatives for function:

a) f(x)= ln(e^-x + 1/x)

2. Use the chain rule. Here is little help to begin with:

$ln(e^{-x} + 1/x)=ln(u)'u'=(1/u)u'=\frac{1}{e^{-x} + 1/x}u'$

Now just find $u'=(e^{-x} + 1/x)'=?$.

Regards.

3. So it would be = $\frac{1}{e^-x + \frac{1}{x}}*(e^-x(-1)-\frac{1}{x^2})$ ???

4. Originally Posted by Stud778
So it would be = $\frac{1}{e^{-x} + \frac{1}{x}}*(-e^{-x}-\frac{1}{x^2})$ ???
Your LaTeX needs work, but that looks right to me.

5. Something is wrong, because the correct answer is

$\frac{-(x-1)}{e^x+x}-\frac{1}{x}$

6. a
Originally Posted by shenanigans87
Something is wrong, because the correct answer is

$\frac{-(x-1)}{e^x+x}-\frac{1}{x}$
$\frac{-(x-1)}{e^x+x}-\frac{1}{x}$

$\frac{-x^2+x-x-e^x}{xe^x+x^2}$

$\frac{-x^2-e^x}{xe^x+x^2}$

$\frac{ \frac{-x^2-e^x}{x^2e^x}}{ \frac{xe^x+x^2}{x^2e^x}}$

$\frac{ - \frac{1}{e^x} - \frac{1}{x^2}}{ \frac{1}{e^x} + \frac{1}{x}}$

$\frac{ -e^{-x} - \frac{1}{x^2}}{ e^{-x} + \frac{1}{x}}$

This argument follows a lot easier if you read it backwards, though.

I kind of like that simplification, though.